Exam 2 Practice Problems - University of Alabama

[Pages:16]PH 101 LeClair

UNIVERSITY OF ALABAMA Department of Physics and Astronomy

Summer 2011

Exam 2 Practice Problems

1. A solid sphere of mass M and radius R starts from rest at the top of an inclined plane (height h, angle

), and rolls down without slipping. What is the linear velocity of the center of mass at the bottom of

the

incline?

For

a

solid

sphere,

I

=

2 5

MR2.

2. A star rotates with a period of 30 days about an axis through its center. After the star undergoes a

supernova explosion, the stellar core, which had a radius of 1.0 ? 104 km, collapses into a neutron star of

radius

3.0

km.

Determine

the

period

of

rotation

of

the

neutron

star.

Note

that

=

2 period

3. A pendulum is made from a rigid rod of length L and mass M hanging from a frictionless pivot point, as shown below. The rod is released from a horizontal position. How does the (tangential) acceleration of the end of the rod at the moment of release compare to g?

L

4. A wheel rotates with a constant angular acceleration = 3.50 rad/s2. If the angular speed is i = 2.00 rad/s at time ti = 0, through what angular displacement does the wheel rotate in 2.00 s? 5. A pendulum consists of a sphere of mass m attached to a light cord of length L. The sphere is released from rest at an angle i from the vertical. Find the speed of the mass at its lowest point. 6. Consider the setup below with two springs connected to a mass on a frictionless table. Find an expression for the potential energy as a function of the displacement x. The springs are at their equilibrium length L when vertical. (Hint: consider the limiting cases L 0 and x 0 to check your solution.)

k L

x L

k

7. A block having a mass m = 0.80 kg is given an initial velocity of v = 1.2 m/s to the right, and it collides with a spring of negligible mass and force constant k = 50 N/m, as shown below. Assuming the surface

to be frictionless, what is the maximum compression of the spring after the collision?

x=0 k

m-v

8. Two blocks are free to slide along a frictionless wooden track ABC as shown in below. The block of mass m1 = 4.92 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

A m1

5.00 m

m2

B

C

9. As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

l

-m v

M

-12 v

10. A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially un-stretched and with force constant k = 1.90 ? 104 N/m, as shown

below. The cannon fires a 200 kg projectile at a velocity of 125 m/s directed 45.0 above the horizontal.

If the mass of the cannon and its carriage is 4780 kg, find the maximum extension of the spring.

45.0

11. Determine the acceleration of the center of mass of a uniform solid disk rolling down an incline making angle with the horizontal.

12. A boy is initially seated on the top of a hemispherical ice mound of radius R. He begins to slide down the ice, with a negligible initial speed. Approximate the ice as being frictionless. At what height does the boy loose contact with the ice?

R

13. A solid sphere of mass m and radius r rolls without slipping along the track shown below. It starts

from rest with the lowest point of the sphere at a height h above the bottom of the loop of radius R,

much larger than r. What is the minimum value of h (in terms of R) such that the sphere completes the

loop?

The

moment

of

inertia

for

a

solid

sphere

is

I

=

2 5

mr2.

m

h

R

Solutions

Solution 1: The simplest approach here is probably conservation of energy. At the top of the ramp, we have only potential energy, while at the bottom we have rotational kinetic energy and translational kinetic energy due to the overall motion of the center of mass (cm).

The vertical height of the ramp is h, and we will presume that the sphere starts out at the very top such that its change in center of mass height is h, rather than h-R. Conservation of energy gives:

Mgh

=

1 2

Mv2cm

+

1 I2 2

(1)

For pure rolling motion, the center of mass velocity must be the same as the velocity of the surface of the sphere, R = vcm, or = vcm/R.

Mgh

=

1 2

Mv2cm

+

1 I

2

v2cm R2

=

1 2

v2cm

I M + R2

(2)

1 2

v2cm

=

Mgh M + I/R2

=

gh 1 + I/MR2

(3)

v2cm

=

1

2gh + I/MR2

(4)

2gh

vcm = 1 + I/MR2

(5)

Using

I

=

2 5

MR2

,

we find

vcm

=

10 7

gh,

slower

than

what

we

would

expect

for

pure

sliding

motion

without friction (v = 2gh).

Solution 2: Presumably, the star's mass is essentially constant. We can use the initial period of the star Ti to find its original rate of rotation i. WIth that in hand, conservation of angular momentum gives use the angular velocity after the explosion, and thus the new period.

2

i = Ti

(6)

Noting that v = R, conservation of angular momentum is straightforward:

Li = mviRi = Lf = mvfRf

(7)

miR2i = mfR2f

(8)

f

=

R2i R2f

i

(9)

Finally, the post-explosion period Tf:

Tf

=

2 f

=

2R2f iR2i

=

Ti

R2f R2i

0.23 s

(10)

Solution 3: If we can get the angular acceleration, we can get the linear acceleration at any point, including the end of the rod. We can get that by finding the net torque on the system. First we should make a quick sketch of the situation:

L/2

mg

The weight of the rod mg acts downward from its center of mass. Since the center of mass is a distance L/s from the pivot point, this will create a torque. The angle between the force mg and a line from the force to the pivot point is 90 when the rod starts to fall, so the magnitude of the torque is just

L

weight = -mg 2

(11)

Here the minus sign indicates that the torque would cause a clockwise rotation, which is by convention

negative. Since this is the only torque present, it must be equal to the moment of inertia times the

angular acceleration. For a thin rod of mass M and length L, rotated about its end, the moment of inertia

is

I=

1 3

ML2.

Thus,

= I

(12)

1 - mgL

=

1 mL2

(13)

2

3

3g

=-

(14)

2L

The linear acceleration a can be found by noting a = R, where R is the distance from the point of rotation to the point of interest, in this case just L:

3

a = -L = - g

(15)

2

The acceleration is half again as large as that due to gravity, meaning the end of the rod falls faster than if

the

rod

were

simply

dropped!

In

fact,

one

can

see

that

a=g

when

R=

2 3

L,

so

points

of

the

rod

further

than that from the pivot point fall faster than dropped objects.

Solution 4: We only need the equation for angle as a function of time under constant angular acceleration:

(t)

=

o

+

it

+

1 t2 2

(16)

In this case, we only want the angular displacement (t) = (t)-o:

(t)

=

it

+

1 t2 2

(17)

We are given i = 2.00 rad/s and = 3.5 rad/s2, so finding the angular displacement after 2.00 s is just a matter of plugging in the numbers:

(2 s) =

rad 2.00

s

(2

s)

+

1 2

rad 3.5 s2

(2 s)2 = 11.0 rad 1.75 rev

(18)

Solution 5: This is a standard conservation of energy problem. Our initial state is the pendulum bob at an angle i, the final state is that where the bob is hanging straight up and down.

L cos

L

L-L cos

The initial energy is purely gravitational potential energy. Geometry gives us the change in height after inclining the bob by i. For convenience we choose the state of zero potential energy to be the final state, with the bob completely vertical. The final energy is purely kinetic. Setting the two equal and solving for v gives us our answer.

Ki + Ui = Kf + Uf

0 + mg(L - L cos i)

=

1 mv2 + 0 2

?m?g(L - L cos i) = 21?m?v2

2gL(1 - cos i) = v2

v = 2gL(1 - cos i)

Solution 6: Since potential energy is a scalar, we can just add the potential energies for each spring together. Since the two springs are equivalent, we can just figure out the potential energy of one of them and double it.

When x = 0, both springs have a length L. As soon as we pull on the mass and move it to some x = 0, we can find the new length of the spring from simple geometry as x2 + L2. The x is the difference between these two lengths. Now we can easily write down PE(x):

PE(x) = PEspring1 + PEspring2

=

2PEspring1 = 2

1 k (x)2 2

= k (x)2

2

= k x2 + L2 - L = kL2 + kx2 - 2kl x2 + L2

= kx2 + 2kL L - L2 + x2

Solution 7: Conservation of energy again. The initial kinetic energy of the block is converted into potential energy stored in the spring.

Ki + Ui = Kf + Uf

1 2

mv2A

+

0

=

0

+

1 k

(x)2

2

1 ? 2?

mv2A

=

1 ? ?

k

(x)2

2

? (x)2

=

?m k

v2A

x =

m

0.8 kg

k vA =

(1.2 m/s) 0.15 m 50 N/m

Solution 8: This one has three basic steps. First, use conservation of energy to find the velocity of the first block just before the collision. Call the initial height h = 5 m, and let the zero for potential energy be the vertical position of m2:

Ki + Ui = Kf + Uf

0 + mgh

=

1 2

mv21i

?m?gh = 21?m?v21i

v1i = 2gh

Next, we handle the collision itself. It is an elastic collision, so both kinetic energy and momentum are conserved. We have already derived the equation for v1f for a 1D elastic collision with one of the objects at rest, and only state the result below.

v1f =

m1 - m2 m1 + m2

v1i =

m1 - m2 m1 + m2

2gh

The last step is to use conservation of energy again to find the new height, which we'll call h .

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